The product of all solutions of the equation $$ e^{5(\log_e x)^{2}+3}=x^{8},x>0 $$, is :

Let $$y=\ln\ x$$

Since $$x>0$$, this substitution is valid.

$$\therefore$$ $$x=e^y$$

and $$x^8=\left(e^y\right)^8=e^{8y}$$

Substitute into the equation 

$$e^{5\left(\ln\ x\right)^2-3}=e^{8y}$$

Since $$\left(\ln\ x\right)=y$$

$$\therefore$$ $$e^{5\left(y\right)^2-3}=e^{8y}$$

Equate the exponents as the bases are same (e)

$$5y^2-3=8y$$

$$\therefore$$ $$5y^2-8y-3=0$$

Sum of the quadratic equation =  $$y_1+y_2$$ which is equal to $$-\dfrac{b}{a}$$

$$\therefore$$ $$y_1+y_2=8/5$$------(i)

To find: $$x_1\times\ x_2$$

We know  $$x=e^y$$ 

$$\therefore$$ $$x_1\times\ x_2=e^{y_1+y_2}$$

Substitute the value of  $$y_1+y_2$$ from---(i)

$$\therefore$$  $$x_1\times\ x_2=e^\dfrac{8}{5}$$.

Hence, option A is the correct answer